Problem: What is the slope of the line tangent to $f(x) = -x^{2}-3x+6$ at $x = 1$ ?
Answer: The slope of the tangent line is $ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-(x+\Delta x)^{2}-3(x+\Delta x)+6) - (-x^{2}-3x+6)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-(x^{2}+2x \Delta x+\Delta x^{2})-3(x+\Delta x)+6) - (-x^{2}-3x+6)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-x^{2}-2(x \Delta x)-\Delta x^{2}-3x-3(\Delta x)+6+x^{2}+3x-6}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-2(x \Delta x)-\Delta x^{2}-3(\Delta x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} -2x-\Delta x-3$ $ = -2x-3$ $ = (-2)(1)-3$ $ = -5$